Algebra (all content)
Course: Algebra (all content) > Unit 11Lesson 20: Advanced interpretation of exponential models (Algebra 2 level)
- Interpreting change in exponential models: with manipulation
- Interpret change in exponential models: with manipulation
- Interpreting change in exponential models: changing units
- Interpret change in exponential models: changing units
- Structure in exponential expression
Structure in exponential expression
Analyzing an elaborate exponential function to determine its value at t=0 and whether it increases or decreases with t. This is algebraic reasoning at its best! Created by Sal Khan.
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- 5:48. How do we know that the expression is going to be always larger than the initial position (c^2 - 1)? I understand that the whole expression will be increasing as (-d^-t) gets smaller as t increases. But how do we make sure that the expression doesn't initially decrease from the starting point and then starts to increase to its maximum? In other words, between (c^2 - 1) and (-d^-t + c^4)/(c^2 + 1), how do we prove that the latter expression is ALWAYS going to be larger than the former?(11 votes)
- Good question! Here is why:
t is always going to be a positive number, because we have defined it as time, and we started our time at 0.
That means that -t will always be negative.
For t>0, d^-t=1/(d^t).
Since d>1, d^t will always be >1 for any t>0.
Therefore, 1/d^t will always be between 0 and 1 and -1/(d^t) will always be between -1 and 0.
So for any t>0, the formula looks like this:
(a small negative number between -1 and 0) + c^4 / (c^2+1).
Then we need to prove that that expression above is always > c^2-1.
I'm going to rename that "small negative number between -1 and 0" to something easier. Let's call it s. -1<s<0.
The formula rewritten using s becomes (s+c^4)/(c^2+1).
Because of the boundaries we set on s, c^4+s > c^4-1. Now let's divide both sides of that inequality by c^2+1.
We have already discovered that we can use difference of squares to make the right-hand side of that equation simplify to c^2-1.
So now we have (c^4+s)/(c^2+1)> c^2-1, which is what we set out to prove!(16 votes)
- why is this video even here?it has nothing related to reasoning about expressions.even if it does , why does the difficulty suddenly increase so much?(12 votes)
- I think (and hope) in some time the KA team will put it in a more advanced section where it belongs.(3 votes)
- Indeed, I don't understand what position of particle is, any help please ?(6 votes)
- Its just a hypothetical scenario to give some context to the equation and make you realise that you could use the same mathematical principles when doing an equation that involves real world things like particles or other physical properties.(6 votes)
- I don´t quite get how does c^4 - 1 is assumed to be = to (c^2)^2 - 1^2 (the -1^2 part gets me confused).
Can anyone explain me that part a little better?(4 votes)
- Sal is trying to get c^4-1 in the a^2-b^2 form, which we know how to factor: a^2 - b^2 = (a + b) (a - b).
1 is equal to 1^2 (which is just 1), so can be used as the b in that equation, that's all Sal's saying.
c^4 - 1 = (c^2)^2 - 1^2 = (c^2+1) (c^2-1)(5 votes)
- How are you supposed to know whether to take 'negative d raised to the t power' or 'd raise to the t power then take the negative'? Shouldn't there be parentheses like -(d^t) not -d^t?(5 votes)
- Exponentiation takes precedence over the minus. I don't know whether you use the PEMDAS mnemonic or not, but the E comes before the S.
So -d^t = -(d^t) = "d raise to the t power then take the negative".
If you want "negative d raised to the t power", then you'd write (-d)^t. The parentheses showing that you want to do the negation first.(3 votes)
- I think I'm missing the big picture here. What are we doing during this video with the function? Thanks.(4 votes)
- We are showing how to visualise function using logic and a few of the tools we have learned throughout Algebra. If you get this stuff then you have probably mastered most of Algebra II. It also leads towards limits which is in precalculus and beyond.(2 votes)
- One of the viewers asked below, "What is a particle?" Well, in this video, Sal is not trying to find out the position of a car or train or plane at a certain time. If he were, the variables would be distance = rate x time or time = distance / rate (divided by rate). Instead, Sal is asking about the position of a particle at a certain time. "Particle" is a term used in physics and it refers to subatomic "particles". These are pieces of matter such as electrons, protons, neutrons, photons and quarks (the substance that makes up protons and neutrons). Think very, very, very small and moving very, very, very, very fast. So, look at the formula in the video again. We have a d-term (distance), a t-term (time) and a c-term (rate). In physics c is the speed of light (approximately 3 x 10^8 meters/second). So, we do have the same terms and the same formula that I started with above re cars, trains and planes, only our rate is in terms of the speed of light instead of mph or kmph. So, Sal is asking what is the location of a subatomic particle (take your pick , let's say a photon) that is moving somewhere close to the speed of light after so many seconds or minutes or years?
In regard to the second question that Sal asks in the video, "What happens as time increases ("t" gets larger)?", I think it is a much more straight forward explanation (easier to comprehend), if you start by simplifying the negative exponent (-t).
[To better follow this explanation, I encourage you to copy the original formula (p(t) = -d^-t +c^4 / c^2 +1 on a piece of paper and manipulate the terms as per my explanation.]
To do this just take the reciprocal of -d^-t. This has the result of putting the term -d^(-t) in the denominator as (-d^t) and (c^4 +1) in the numerator. Now, (c^4 + 1) in the numerator is a constant (doesn't change over time). And in the denominator the term (c^2 + 1) is also a constant (doesn't change over time). Since the minus d^t (we got rid of the -t) in the denominator will increase exponentially (think very quickly) over time and this increasing value is being subtracted from the constant term in the denominator (c^2 +1), the result is the denominator will be get smaller and smaller (approaching 0) as time gets larger (the passage of time). We, all know that for a fraction as the denominator gets smaller the number represented by the fraction gets larger. So, the answer is: for p(t), as t gets larger the location of this subatomic particle will be some location further and further away (will get larger). This makes sense. Think of a car moving away from your house at a fixed speed traveling down a straight highway. At any given time, you could tell its location (let's say on a map) after so many minutes or hours. And, this location would be further and further away from your house as time elapsed.
Another solution to this equation is, once again starting with the original equation and simplify the negative exponent (-t) by taking the reciprocal of (-d^-t). This gives us (-1/d^t). So, we have a negative fraction (meaning either the numerator is negative or the denominator is negative). This time let's make the numerator negative which gives us: p(t) = (C^4-1) / (c^2 +1) +d^t. Now, as Sal shows us in the video, (c^4-1) = (c^2+1)(c^2-1) [the difference of 2 squares - this topic is taking up in algebra a little bit later). But, if you multiplied those 2 factors, you would end up with (c^4-1). Now, rewrite the equation as p(t) = (c^2+1)(c^2-1) / (c^2+1) + d^t. As, you see the (c^2+1) in the numerator cancels the (c^2+1) in the denominator leaving: p(t) = (c^2-1) / d^t. Once again, the numerator is a fixed constant while the denominator gets larger and larger as time (t) increases. This leaves the same result as in the first explanation (above): as the denominator increases the number represented by the fraction gets larger.(4 votes)
- What does c^2-1 represent? when time is 0 its position is at c^2 - 1 distance?(2 votes)
- I am trying to listen to this and make sense of it. It has been 40 years since I studied mathematics in school and even though I have an MA in something else I can't make head of tails of this. Why is -d to the -t equal to 0? I don't remember ever learning about 0 exponents or negative exponents. I do remember that -d to the 2nd would be -d*-d. So I suppose that -d to the 1st would just be plain -d. But why does d to the 0 become 1 and not 0?(2 votes)
- Exponents follow a rule: (x^a)*(x^b)= x^(a+b)
If you plug some positive whole numbers into this as a and b, you see that this gels with our notion of exponents as repeated multiplication. So let's see what happens with 0.
x^0 = (x^a)/(x^a)
If you like, Ben Orlin (link) gives a nice explanation of this, plus a bit more, from the standpoint of a teacher.
- Can anyone help me with the next practice question, I'm flummoxed. The volume (in cubic meters), v, of a rectangular room is given by the expression:
v = 162-2b^6
Where b is a positive integer and each dimension is an integer greater than 1 meter. What are three unique expressions that could represent the dimensions of the room in terms of b?
In the answer box, write these three expressions as a product that equals the volume (e.g. "(expression 1)(expression 2)(expression 3)").
To be perfectly honest I don't even understand the question. But I came up with:
Thus my three unique expressions when all multiplied together add up to v but it's not quite there. Any advice or assistance would be great.(2 votes)
- You want three expressions whose product equals the volume. Fortunately, the expression 162-2b^2 can be factored into 3 terms:
First factor out a 2:
2(81 - b^6)
Then use difference of squares:
2(9^2-(b^3)^2) = 2(9+b^3)(9-b^3)
Thus the 3 expressions are: (2), (9+b^3), (9-b^3)(3 votes)
- I think I have missed the entire philosophy and driving manual for Khanacademy. I now have 4 blue boxes on my member entry screen. All of them have to do with algebra but after looking at this last string of tutorials about structure in exponential expression I BELIEVE I HAVE SOMEHOW MOVED INTO ALGEBRA 2. At least I hope so, because this makes no sense and the leap in difficulty from last week to this week is frustrating and makes coming back to learn a P.I.A. Does a new learning box appear every time I attempt to go back and go over something again? How do I even know right now where I am supposed to be in learning algebra I? I was having troubles with functions and suddenly got to a mastering review page. Final comment: How do I start over?(3 votes)
- Go to this page: https://www.khanacademy.org/mission/algebra
On the left hand side you will see a progress meter. Just below that is a link that says "Show All Skills". Click on that to see what you have done and what is left to do.
Hover over each item in the skills detail to see what the skill is. Click on it if you want to practice it again.(0 votes)
Let's say that the position of some particle as a function of time is given by this expression right over here. Negative d to the negative t power plus c to the fourth over c squared plus 1, where c and d are constants and both of them are greater than one. So what I want to do over the course of this video is see what can we infer based on this expression, this function definition, that we have here. And the first thing that I want you to think about is, what is the initial position? If I were to express the initial position, in terms of c's and d's, and try to simplify it. So I encourage you to pause the video and try to find an expression for the initial position. Well, the initial position is the position we're at when our time is equal to 0. So we essentially just want to find p of 0. And p of 0 is going to be equal to negative d to the negative 0. So I'll just write that down. Negative 0 plus c to the fourth over c squared plus 1. Well, d to the negative 0, that's the same thing as d to the 0, and since we know that d is non-zero, we know this is defined. Anything non-zero to the 0-th power is going to be 1. And the zero is actually under debate, what 0 to the 0-th power is. But we can safely say that this right over here is going to be equal to 1. And so the numerator here simplifies 2. This is equal to-- and I'll switch the order. c to the fourth minus 1 over c squared plus 1. And now this might jump out at you as a difference of squares. We could write this as c squared, squared, minus 1 squared over c squared plus 1. And that's the same thing as c squared plus 1 times c squared minus 1. All of that over c squared plus 1. And we have a c squared plus 1 in the numerator and that denominator so we can simplify. And so our initial position is going to be c squared minus 1. So that actually simplified out quite nicely. Now the next question I'm going to ask you is-- OK, we know that the initial position that time equals zero, the particle is going to be at c squared minus 1. But what happens after that? Does the position keep increasing? Does the position keep decreasing? Or does the position maybe increase and then decrease, or decrease and then increase as in keep swapping around? So I encourage you to pause the video now and think about what happens to the position. Does it keep increasing? Does it keep decreasing? Or does it do something else? Well, let's answer that question of what's happening to the position after our initial position. We really just have to focus on this term right over here. This d to the negative t. This is the only part that really t is driving. All these other things are staying the same as we go through time. So what happens to d to the negative t power, to this part of this first term, as t goes from 0 onwards. And to think about that, let's plot. Let's plot what the function d to the negative t looks like. d to the negative t would look like-- and we know that d is greater than 1. So when t is equal to 0-- So this is t right over here. And over here we're going to plot on this axis. We're going to plot d to the negative t. When t is equal to 0 this is going to be equal to 1. We've already seen that. Now what happens as t increases? Say t increases to 1. Well, now this is going to be d to the negative 1, which is the same thing as 1 over d. And we don't know the exact value for d, but we know, since d is greater than 1, 1 over d is going to be less than 1. So let's say this is 1 over d right over here. 1 over d. So it's going to be something like that. And then when t is at 2 we're going to be at 1 over d squared, which is going to be something like right over there. And you see at least this term what it's doing as t increases. As t increases, d to the negative t is strictly decreasing. And once again we know that because d is greater than 1. So this term right over here is strictly decreasing. So this is decreasing. That part of it. But we're not adding it. We're subtracting it. We're subtracting from the beginning. At first this starts off at 1. We subtract 1. And then we start subtracting smaller and smaller things than 1. So if this is decreasing, but we're subtracting it, we're subtracting smaller and smaller things, this whole thing, the negative of it, is going to be increasing. Another way to think about it. If you wanted to plot negative d to the negative t, it would look like this. It would just be the negative of this right over here. So it would look something like-- I'll do that yellow color-- this. So this whole term right over here, negative d to the negative t, is constantly increasing. And we know that all of these other things, well, these are just going to be fixed. So this entire expression is constantly increasing starting at t equals 0. And then t going into larger and larger positive values. Now the last thing I want to ask you is, what is the maximum value here? What is a value that this will never be able to get to? It might try to get close to it, but it's never going to quite get there. Well, we already know that it's increasing, but let's just think about what happens as t becomes really, really, really large numbers. Really you could think about it as t approaches infinity. Well, once again let's look at this d to the negative t term. You see the d to the negative t, as t gets larger and larger, is getting smaller and smaller. This term right over here is approaching 0 as t goes to infinity. Well, if this is approaching 0 that means we're subtracting 0. So this whole yellow thing, the negative d to the negative t, is increasing but it's increasing at a smaller and slower rate. So this is the negative d to the negative t. This is right over here. You see it increases but it never quite gets to the horizontal axis right over here. And so if we think about it as t approaches infinity, this whole thing just becomes 0 and our entire position is approaching but never quite gets to c to the fourth over c squared plus 1. So one way to think about it, it's approaching this but it never quite gets to a position of c to the fourth over c squared plus 1.